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\begin{document}

\title{高等代数二}
\subtitle{01-一元多项式环-多项式的整除性 }
%\institute{上海立信会计金融学院}
%\author{王立庆}
\author{{\ppr LQW}}
\renewcommand{\today}{{\ppr \number\year \,年 \number\month \,月 \number\day \,日} }
%\date{{\ppr 2023年2月28日} }

\maketitle

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\begin{frame}{1.1. 作业：星期天晚上十点半之前在网络教学平台提交 }

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}
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\begin{enumerate}
\item   整理课堂笔记里的重点难点。补充没写完的计算或证明。
\item   习题(2.1)任选2题。抄写题目。
\item   习题(2.2)任选2题。抄写题目。
\end{enumerate}

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\begin{enumerate}

\item   数环
\item   数环上的一元多项式
\item   一元多项式的加法、减法、乘法
\item   一元多项式的次数
\item   一元多项式环
\item   数域
\item   数域上的一元多项式
\item   多项式整除的基本性质
\item   一元多项式的除法
\item   带余除法定理

\end{enumerate}

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\begin{enumerate}

\item   举例说明多项式的长除法与综合除法。
\item   证明带余除法的存在性与唯一性。
\item   使用多项式的次数进行证明。
\item   证明多项式的整除的基本性质。
\item   理解数环与数域的概念，理解一元多项式环的概念。

\end{enumerate}

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\begin{itemize}


\item  {\color{red}问：什么是数环？}

\item  答：数环是一些复数组成的集合，含有0和1，并且这个集合中的任意两个元素在加法、减法和乘法运算下的结果仍然在这个集合里。例如整数全体是一个数环。


\item {\color{red}问：什么是数环 $R$ 上的一元多项式？}

\item  答：是指这样一个表达式 $$a_0+a_1x+a_2x^2+\cdots+a_nx^n,$$
其中 $x$ 是一个字母，$a_0,a_1,\cdots,a_n$ 是数环 $R$ 中的元素，称为系数。

\end{itemize}

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\begin{itemize}

\item  {\color{red}问题：下述哪些数集是数环？} 
\begin{enumerate}
\item  {\color{red}整数全体组成的集合：$\mathbb{Z}$. } 
\item  {\color{red}有理数全体组成的集合：$\mathbb{Q}$. } 
\item  {\color{red}实数全体组成的集合：$\mathbb{R}$. } 
\item  {\color{red}复数全体组成的集合：$\mathbb{C}$. } 
\item  {\color{red}高斯整数全体组成的集合：$\mathbb{Z}[i] = \{m+ni\mid m,n\in\mathbb{Z}\}$. } 
\end{enumerate}

\item  解答：ABCDE. 


\end{itemize}

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\begin{itemize}

\item  {\color{red}问：什么时候称两个多项式相等？}

\item  答：当这两个多项式的系数都对应相等的时候。

\item  {\color{red}问：什么是一个多项式的次数？}

\item  答：设多项式 $f(x)=a_0+a_1x+a_2x^2+\cdots+a_nx^n,$ 其中 $a_n\neq 0$, 则称这个多项式的次数为 $n$, 记为
$$\deg (f(x))=n.$$

\item  注：零多项式不定义次数。（有的书定义零多项式的次数为负无穷大。）

\end{itemize}

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\begin{itemize}

\item  {\color{red}问：两个多项式的和、差、积是怎样定义的？}

\item  答：按照最直观的方式定义。和、差分别是系数对应相加、相减。乘法是按分配律两两相乘，然后合并同次项的系数。

\item  {\color{red}问：什么是数环 $R$ 上的字母 $x$ 的一元多项式环？}

\item  答：是指以数环 $R$ 中的数作为系数的所有多项式组成的集合，记为 $$R[x].$$ 

\vfill 

\item  注：集合 $R[x]$ 中的两个多项式在加法、减法和乘法运算下的结果仍然在集合 $R[x]$ 中。具有这种性质的集合连同运算，即 $$(R[x], +, \cdot),$$ 称为是一个``环''，这是一种代数系统。

\end{itemize}

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\begin{itemize}

%\item  {\color{red}问：环是什么？}

\item  {\color{red}定义：一个环是一种代数结构，是指这样一个三元组 $$(S,+,\cdot),$$
其中 $S$ 是一个集合，还有两种运算，
\begin{eqnarray*}
+: && S\times S \to S, \\
\cdot: && S\times S\to S,
\end{eqnarray*}
它们之间满足结合律、分配律、交换律等运算规律。}

\end{itemize}

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\begin{itemize}

\item  {\color{red}定理2.1.1. }
\begin{itemize}
\item  {\color{red} 两个多项式的和多项式的次数，不会超过这两个多项式的次数的较大值。}
\item  {\color{red} 两个多项式的积多项式的次数，等于这两个多项式的次数的和。}
\end{itemize}


\item  证明：
\begin{enumerate}
\item  将这两个多项式的具体形式写出来。
\item  按加法计算，最高次项可能会相互抵消，所以次数不会超过。
\item  按乘法计算。各自的最高次项相乘，是乘积的最高次项。
\item  按次数的定义证明。关键是两个非零复数的乘积不会等于零。
\end{enumerate}

\end{itemize}

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\begin{itemize}

\item  {\color{red}推论2.1.1. 两个多项式的乘积如果是零多项式，那么其中至少有一个是零多项式。}

\item  证明：按照多项式乘积的次数来论证。

\vspace{1cm}

\item  {\color{red}推论2.1.2. 多项式乘积的消去律成立。即：如果 $f(x)g(x)=f(x)h(x)$ 且 $f(x)\neq 0$, 那么有 $g(x)=h(x)$. }

\item  证明：由条件得 $f(x)[g(x)-h(x)]=0$. 因 $f(x)\neq 0$, 由推论2.1.1 得证。

\end{itemize}

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\begin{itemize}

\item  {\color{red}问：什么是数域？}

\item  答：数域是一种特殊的数环，当除数不为零的时候，除法的结果也在这个数环里。例如整数集是数环但不是数域，有理数集既是数环也是数域。实数集也既是数环也是数域。


\end{itemize}

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\begin{itemize}

\item  {\color{red}问题：下述哪些数集是数域？}
\begin{enumerate}
\item  {\color{red}整数全体组成的集合：$\mathbb{Z}$. }
\item  {\color{red}有理数全体组成的集合：$\mathbb{Q}$. }
\item  {\color{red}实数全体组成的集合：$\mathbb{R}$. }
\item  {\color{red}复数全体组成的集合：$\mathbb{C}$. }
\item  {\color{red}高斯整数全体组成的集合：$\mathbb{Z}[i] = \{m+ni\mid m,n\in\mathbb{Z}\}$. }
\end{enumerate}

\item  解答：BCD. 


\end{itemize}

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\begin{itemize}

\item  {\color{red}问：什么是多项式的整除？}

\item  答：设 $F$ 是一个数域，设 $f(x)$ 和 $g(x)$ 是系数在 $F$ 中的两个多项式。如果存在系数在 $F$ 中的多项式 $h(x)$ 使得 $$f(x)=g(x)h(x),$$ 那么我们称 $g(x)$ 整除 $f(x)$, 并用符号 $g(x)\mid f(x)$ 表示。

\item  例子：$x-1$ 整除 $x^2-1$, 这个事实可以写成 $(x-1)\mid (x^2-1)$. 

\end{itemize}

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\begin{itemize}

\item  {\color{red}问：写出整除的一些基本性质，并加以证明。}

\item  答：按照整除的定义来证明。

\begin{enumerate}
\item  若 $f\mid g$ 且 $g\mid h$, 则 $f\mid h$. 
\item  若 $h\mid f$ 且 $h\mid g$, 则 $h\mid (f\pm g)$. 
\item  若 $h\mid f$, 则对任意 $g\in F[x]$, 有 $h\mid fg$. 
\item  若 $h\mid f_i \,\,(\forall i=1,2,\cdots, t)$, 则对任意 $g_i\in F[x]$, 有 $h\mid \sum_{i=1}^t f_ig_i$. 
\item  对任意 $c\in F, c\neq 0$, 对任意 $f\in F[x]$, 都有 $c\mid f$. 
\item  对任意 $c\in F, c\neq 0$, 对任意 $f\in F[x]$, 都有  $cf\mid f$ 与 $f\mid cf$.  
\item  若 $f\mid g$ 且 $g\mid f$, 则存在 $c\in F,c\neq 0$ 使得 $f=cg$.    
\end{enumerate}

\end{itemize}

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\begin{itemize}

\item  {\color{red}问：什么是带余除法？}

\item  答：如果 $f(x)=g(x)q(x)+r(x)$, 而且 $$r(x)=0 \text{ 或者 } \deg(r(x)) < \deg (g(x)),$$ 
那么称 $f(x)$ 除以 $g(x)$ 的商是 $q(x)$, 余式是 $r(x)$. 

\end{itemize}

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\begin{itemize}

\item  {\color{red}问题：整数环和一元多项式环中的带余除法举例。}

\item  解答：
\begin{eqnarray*}
2023 \div 20 &=& 101 \cdots 3 \\ 
(x^2+1) \div (x+1) &=& (x-1) \cdots 2 \\
(x^3+3) \div (x^2-1) &=& x \cdots (x+3)
\end{eqnarray*}


 \item  注：带余除法是两个多项式不能整除的时候的处理方法。
 
\end{itemize}

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\includegraphics[height=0.7\textheight, width=0.9\textwidth]{pic/synthetic-division.jpg}
\end{center}

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\begin{itemize}

\item  {\color{red}定理2.2.1. 在数域 $F$ 上的一元多项式环中，带余除法的结果是存在且唯一确定的。即设 $f(x),g(x)\in F[x]$, 则存在唯一的一对多项式 $q(x)$ 与 $r(x)$, 使得
$f(x)=g(x)q(x)+r(x)$, 且 $r(x)=0$ 或 $\deg r(x) < \deg g(x)$. 
}

\item  证明：

\begin{enumerate}
\item  存在性：要证明符合这个等式的 $q(x)$ 与 $r(x)$ 存在。对 $f(x)$ 的次数进行归纳法证明。
\item  唯一性：设有下述两个式子成立，
\begin{eqnarray*}
f(x)=g(x)q_1(x)+r_1(x)  \text{ 且 }\, r_1(x)=0 \text{ 或 }\, \deg(r_1(x))< \deg(g(x)),\\
f(x)=g(x)q_2(x)+r_2(x)  \text{ 且 }\, r_2(x)=0 \text{ 或 }\, \deg(r_2(x))< \deg(g(x)).
\end{eqnarray*}
要证明 $g_1(x)=g_2(x)$, $r_1(x)=r_2(x)$. 
\end{enumerate}

\end{itemize}

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\begin{itemize}

\item  {\color{red}证明带余除法的存在性：}

\begin{enumerate}

\item  固定 $g(x)$ 的次数，设为 $m$. 当 $f(x)$ 的次数小于 $m$ 时可证。
\item  当 $f(x)$ 的次数大于等于 $m$ 时，设 $f(x)$ 的次数为 $n$. 则选取适当的 $c\in F$, 使得
$$f(x)- cx^{n-m}g(x)$$
的次数小于 $n$. 
\item  由归纳假设可得存在 $q_1(x)$ 与 $r_1(x)$ 使得
$$f(x)- cx^{n-m}g(x) = g(x)q_1(x) + r_1(x). $$
 

\end{enumerate}

\end{itemize}

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\begin{itemize}


\item  {\color{red}证明带余除法的唯一性：}

\begin{enumerate}

\item  两个式子的左边相等，所以有
$$g(x)q_1(x)+r_1(x) = g(x)q_2(x)+r_2(x).$$

\item  移项得 
$$g(x) [ q_1(x) -q_2(x) ] = r_2(x) -r_1(x).$$

\item  比较等式两边的次数。

\end{enumerate}


\end{itemize}

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\begin{itemize}

\item  {\color{red}命题1：设 $R$ 是一个数环，设 $f(x),g(x)\in R[x]$, 且 $g(x)$ 的首项系数为1，则存在唯一的一组多项式 $q(x),r(x)\in R[x]$ 使得
带余除法成立，即 $$f(x)=g(x)q(x)+r(x), \,\text{ 且 }\, r(x)=0 \text{ 或 } \deg(r(x))< \deg(g(x)). $$
}

\item  证明思路：检查带余除法定理的证明，系数域 $F$ 是数域这个性质用在哪里。
条件``$g(x)$ 的首项系数为1'' 如何使得这个数域的性质可以不用。

\end{itemize}

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\begin{itemize}


\item  {\color{red}命题2：设有两个有理系数多项式 $f(x)$ 与 $g(x)$, 如果不存在有理系数多项式 $h(x)$ 使得 $f(x)=g(x)h(x)$, 那么也不存在实系数多项式 $h(x)$ 使得 $$f(x)=g(x)h(x). $$
}

\item  证明：
\begin{enumerate}

\item  在有理系数范围内进行带余除法，有 $f(x)=g(x)q(x)+r(x)$. 
\item  由条件知 $r(x)\neq 0$. 
\item  在实系数范围内进行带余除法，由带余除法的唯一性，也一定只有这个等式。
\item  由 $r(x)\neq 0$ 且 $r(x)$ 的次数小于 $g(x)$ 的次数得证。
\end{enumerate}

\end{itemize}

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\begin{itemize}

\item {\color{red}问题：设 $f(x),g(x)$ 和 $h(x)$ 是实数域上的多项式。证明：若
\begin{eqnarray*}
f(x)^2 = xg(x)^2 + xh(x)^2,
%\label{eq-2-1-1}
\end{eqnarray*}
则 $f(x)=g(x)=h(x)=0$.
}

\item  解答：考察最高项系数。


\end{itemize}

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\begin{itemize}

\item  {\color{red}问题：找出一组满足等式
\begin{eqnarray*}
f(x)^2 = xg(x)^2 + xh(x)^2,
%\label{eq-2-1-1}
\end{eqnarray*}
的不全为零的复数系数的多项式 $f(x),g(x)$ 和 $h(x)$.  
}

\item  解答：使用虚数。


\end{itemize}

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\begin{itemize}

\item  {\color{red}问题：证明下述多项式的恒等式，
{\footnotesize 
\begin{eqnarray*}
&& 1-x+\frac{x(x-1)}{2!}-\frac{x(x-1)(x-2)}{3!}+\cdots+(-1)^n\frac{x(x-1)(x-2)\cdots (x-n+1)}{n!} \\ 
&=& (-1)^n\frac{(x-1)(x-2)\cdots (x-n)}{n!}.
%\label{eq-2-1-3}
\end{eqnarray*}
}
}

\item  解答：几种思路：
\begin{enumerate}
\item  分别代入 $x=1,2,\cdots,n$. 验证左边等于零。
\item  对 $n$ 进行数学归纳法。
\end{enumerate}

\end{itemize}

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\begin{itemize}

\item  {\color{red}问题：求多项式 $f(x)$ 除以多项式 $g(x)$ 得到的商式和余式，
\begin{eqnarray*}
f(x) &=& x^4-4x^3-1, \\
g(x) &=& x^2-3x-1.
%\label{eq-2-2-1}
\end{eqnarray*}
}

\item  解答：长除法。商式 $q(x)=x^2-x-2$, 余式 $r(x)=-7x-3$. 


\end{itemize}

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\begin{itemize}

\item  {\color{red}求证：$x\mid f(x)^k$ 当且仅当 $x\mid f(x)$. }

\item  解答：使用带余除法定理。

\end{itemize}

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\begin{itemize}

\item  {\color{red}问题：设 $f_1(x), f_2(x), g_1(x), g_2(x)$ 都是数域 $F$ 上的多项式，其中 $f_1(x)\neq 0$, 且 
$g_1(x)g_2(x)\mid f_1(x)f_2(x)$, $f_1(x)\mid g_1(x)$. 证明：$g_2(x)\mid f_2(x)$. 
}

\item  解答：考虑整除的定义。

\end{itemize}

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\begin{itemize}

\item  {\color{red}问题：实数 $m,p,q$ 满足什么条件时，多项式 $x^2+mx+1$ 能够整除多项式 $x^4+px+q$ ? }

\item  解答：长除法。条件是 $p=m^3-2m$, $q=m^2-1$. 


\end{itemize}

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\begin{itemize}

\item  {\color{red}问题：设 $F$ 是一个数域，$a\in F$. 证明：$x-a$ 整除 $x^n-a^n$. }

\item  解答：验证整除的定义。


\end{itemize}

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\begin{itemize}

\item  {\color{red}问题： 考虑有理系数多项式 
\begin{eqnarray*}
f(x) = (x+1)^{k+n} + (2x)(x+1)^{k+n-1}+\cdots+(2x)^k(x+1)^n, 
%\label{eq-2-2-6}
\end{eqnarray*}
其中 $k$ 和 $n$ 都是非负整数。证明：$$x^{k+1}\mid (x-1)f(x)+(x+1)^{k+n+1}. $$
}

\item  解答：观察可知 $f(x)$ 提出因子 $(x+1)^n$ 后，剩下的部分形如 $$a^k+a^{k-1}b+\cdots+ab^{k-1}+b^k. $$
考虑恒等式
$$(a-b)(a^k+a^{k-1}b+\cdots+ab^{k-1}+b^k) = a^{k+1} - b^{k+1}. $$

\end{itemize}

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\begin{itemize}

\item  {\color{red}问题：证明 $x^d-1$ 整除 $x^n-1$ 当且仅当 $d$ 整除 $n$.  } 

\item  解答：先试试 $x^2-1$ 是否整除 $x^3-1$, $x^4-1$, $x^5-1$. 

\end{itemize}

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